**21.(iii)**** ** In a class of students, opted for NCC, opted for NSS and opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(iii) The student has opted NSS but not NCC.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.
Given,
n(S) = 60, n(A) = 30, n(B) =32, n(A B) = 24
Therefore, P(A) =
P(B) =
P(A B) =
(iii) Now,
Probability that the student has opted NSS but not NCC = P(B A' ) = P(B-A)
We know,
P(B-A) = P(B) - P(A B)
Hence,the probability that the student has opted NSS but not NCC is

**2****1.(ii) ** In a class of students, opted for NCC, opted for NSS and opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(ii) The student has opted neither NCC nor NSS.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.
Given,
n(S) = 60, n(A) = 30, n(B) =32, n(A B) = 24
Therefore, P(A) =
P(B) =
P(A B) =
(ii) Now,
Probability that the student has opted neither NCC nor NSS = P(A' B' )
We know,
P(A' B' ) = 1 - P(A B) [De morgan's law]
And, P(A B) = P(A)+ P(B) - P(A B)
P(A' B' )
Hence,the probability...

**21.(i)** In a class of students, opted for NCC, opted for NSS and opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(i) The student opted for NCC or NSS.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.
Given,
n(S) = 60, n(A) = 30, n(B) =32, n(A B) = 24
Therefore, P(A) =
P(B) =
P(A B) =
(i) We know,
P(A B) = P(A)+ P(B) - P(A B)
Hence,the probability that the student opted for NCC or NSS is

**20.**** **The probability that a student will pass the final examination in both English and Hindi is and the probability of passing neither is . If the probability of passing the English examination is , what is the probability of passing the Hindi examination?

Let A be the event that the student passes English examination and B be the event that the students pass Hindi examination.
Given,
P(A)=0.75, P(A B) = 0.5, P(A' B') =0.1
We know,
P(A' B') = 1 - P(A B)
P(A B) = 1 - 0.1 = 0.9
Also,
P(A B) = P(A)+ P(B) - P(A B)
P(B) = 0.9 - 0.75 + 0.5 = 0.65
Hence,the probability of passing the Hindi examination is 0.65

**19. **In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is and the probability of passing the second examination is . The probability of passing atleast one of them is . What is the probability of passing both?

Let A be the event that the student passes the first examination and B be the event that the students passes the second examination.
P(A B) is probability of passing at least one of the examination.
Therefore,
P(A B) = 0.95 , P(A)=0.8, P(B)=0.7
We know,
P(A B) = P(A)+ P(B) - P(A B)
P(A B) = 0.8 + 0.7 - 0.95 = 1.5 -0.95 = 0.55
Hence,the probability that the student will pass both the...

**18. **In Class XI of a school of the students study Mathematics and study Biology. of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Let M denote the event that the student is studying Mathematics and B denote the event that the student is studying Biology
And total students in the class be 100.
Given, n(M) = 40 P(M) =
n(B) = 30 P(M) =
n(M B) = 10 P(M) =
We know,
P(A B) = P(A)+ P(B) - P(A B)
P(M B) = 0.4 + 0.3 - 0.1 = 0.6
Hence, the probability that he will be studying Mathematics or Biology is 0.6

**17(iii) **A and B are events such that P(A) , P(B) and P(A and B) . Determine (iii) P(A or B)

A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (ii) P(not B)

**17.(ii) **A and B are events such that P(A) , P(B) and P(A and B) . Determine (ii) P(not B)

**17.(i) **A and B are events such that P(A) , P(B) and P(A and B) . Determine (i) P(not A)

Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.

**16. ** Events E and F are such that P(not E or not F) , State whether E and F are mutually exclusive.

Given,
For A and B to be mutually exclusive,
Now,
We know,
Hence, E and F are not mutually exclusive.

**15.(ii) ** If E and F are events such that , and find (ii) P(not E and not F).

**15.(i) ** If E and F are events such that , and , find (i) P(E or F)

**14. ** Given and . Find , if and are mutually exclusive events.

**13 ** Fill in the blanks in following table:

(i) | ||||

(ii) | ||||

(iii) |

**12.(ii) ** Check whether the following probabilities and are consistently defined

(ii)

(ii) Given,
We know,
P(A B) = P(A)+ P(B) - P(A B)
0.8 = 0.5 + 0.4 - P(A B)
P(A B) = 0.9 - 0.8 = 0.1
Therefore, P(A B) < P(A) and P(A B) < P(B) , which satisfies the condition.
Hence, the probabilities are consistently defined

**12.(i) ** Check whether the following probabilities and are consistently defined:

(i)

(i) Given,
Now P(A B) > P(A)
(Since A B is a subset of A, P(A B) cannot be more than P(A))
Therefore, the given probabilities are not consistently defined.

**10.(ii)** A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (ii) a consonant

Given, ‘ASSASSINATION’
No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1
No. of letters = 13
No. of consonants = {4 S's,2 N's,T} = 7
One letter is selected:
n(S) = = 13
Let E be the event of getting a consonant.
n(E) = = 7

**10.(i) ** A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel

Given, ‘ASSASSINATION’
No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1
No. of letters = 13
No. of vowels = {3 A's,2 I's,O} = 6
One letter is selected:
n(S) = = 13
Let E be the event of getting a vowel.
n(E) = = 6

**9.** If is the probability of an event, what is the probability of the event ‘not A’.

**8.(ix)** Three coins are tossed once. Find the probability of getting

(ix) atmost two tails

Sample space when three coins are tossed: [Same as a coin tossed thrice!]
S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]
Let E be the event of getting atmost 2 tails = Event of getting 2 or less tails = {HHT, HTH, THH, TTH, HTT, THT}
n(E) = 6
The required probability of getting atmost 2 tails is .

- Previous
- Next

Exams

Articles

Questions